similar to: Independent contrasts from lme with interactions

Trying to use contrast to look at differences within an lme lme.fnl.REML <- lme(Max ~ S + Tr + Yr + Tr:Yr, random = ~1 |TID, method = "REML") I have three levels of Tr I'm trying to contrast among different years (R, T97, T98), years = 1997-1999, so I'm interested in contrasts of the interaction term. > anova(lme.fnl.REML) numDF denDF F-value

On 5/2/07, r-help-request@stat.math.ethz.ch < r-help-request@stat.math.ethz.ch> wrote: > > Send R-help mailing list submissions to > r-help@stat.math.ethz.ch > > To subscribe or unsubscribe via the World Wide Web, visit > https://stat.ethz.ch/mailman/listinfo/r-help > or, via email, send a message with subject or body 'help' to >

We run ssh from a program and needed to add port-forwards dynamically. The ~C method turns out to be very cumbersome to use since it reads from /dev/tty. But then I came to think of the master/slave functionality (which we already used) which seemed a perfect place for this functionality. Unfortunately it turned out not to be possible to set up new port forwards in a slave. So I patched openssh

Dear R users I have a big matrix like 6021 1188 790 290 1174 1015 1990 6613 6288 100714 6021 1 0.658 0.688 0.474 0.262 0.163 0.137 0.32 0.252 0.206 1188 0.658 1 0.917 0.245 0.331 0.122 0.148 0.194 0.168 0.171 790 0.688 0.917 1 0.243 0.31 0.122 0.15 0.19 0.171 0.174 290 0.474

This command cdmoutcome<- glm(log(value)~factor(year) > +log(gdppcpppconst)+log(gdppcpppconstAII) > +log(co2eemisspc)+log(co2eemisspcAII) > +log(dist) > +fdiboth > +odapartnertohost > +corrupt > +log(infraindex) > +litrate > +africa >

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Dear all, I am so ashamed to pollute the list with a trivial question, but it is a long time I have not used R, and I need a result in the next one or two hour... I have a table which I have loaded with read.table, and I want to make the mean of its columns. > slides <- read.table("slides.txt") > slides [1:5,] V1 V2 V3 V4 V5 V6 V7 V8 1

Hi all, I am dealing with a big data frame. When printing something like > allData[[3]] 1 625.364 38.223 21.014 0.216 1.241411 V 1050o 58.38065 -0.06178768 2 383.709 55.811 21.435 0.296 1.241411 V 1050o 58.38308 -0.03328282 3 434.669 58.597 21.207 0.233 1.241411 V 1050o 58.38334 -0.03930350 4 687.306 69.418 20.873 0.171 1.241411 V 1050o 58.38425 -0.06914694 5 759.522

Hi all, First I dont have much experience with R so be gentle. OK, I am dealing with a dataset (~ tens of thousand lines, each line ~ 10 columns of data). I have to create some subset of this data based on some certain conditions (for example, same first column with another dataset etc...). Here is how I did it: # import data dat <- read.table( "test.txt", header=TRUE,

Hello, I have been getting the following intermittent error from kmeans: >str(cavint.p.r) num [1:1967, 1:13] 0.691 0.123 0.388 0.268 0.485 ... - attr(*, "dimnames")=List of 2 ..$ : chr [1:1967] "6" "49" "87" "102" ... ..$ : chr [1:13] "HYD" "NEG" "POS" "OXY" ... > set.seed(34) >

Dear all, I have an annual time-series of population numbers and I would like to estimate the auto-correlation. Can I use acf() function and judge whether auto-correlation is significant by the plots? The acf array, eg: Autocorrelations of series 'x$log.s.r', by lag 0 1 2 3 4 5 6 7 8 9 10 11 12 1.000 0.031 -0.171

This is probably simple, but I just can't see it... I want to calculate the R^2s for a series of linear models where each term is dropped in turn. I can get the RSS from drop1(), and the r.squared from summary() for a given model, but don't know how to use the result of drop1() to get the r.squared for each model with one term dropped. Working example: library(vcd) # for

Good Morning. Below I like statement like j<-grep(".r\\b",colnames(mydata),value=TRUE); j with the \\b option which I read long time ago which Ive found useful. Are there more or these options, other than ? grep? Thanks. dstat is just my own descriptive routine. > x ?[1] "age"????????? "sleep"??????? "primary"????? "middle" ?[5]

Dear List: I have a data frame prepared in the couting process style for including a binary time-dependent covariate. The first few rows look like this. PtNo Start End Status Imp 1 1 0 608.0 0 0 2 2 0 513.0 0 0 3 2 513 887.0 0 1 4 3 0 57.0 0 0 5 3 57 604.0 0 1 6 4 0 150.0 1 0 The outcome

Dear all, I am using the publically available GustoW dataset. The exact version I am using is available here: https://drive.google.com/open?id=0B4oZ2TQA0PAoUm85UzBFNjZ0Ulk I would like to produce a nomogram for 5 covariates - AGE, HYP, KILLIP, HRT and ANT. I have successfully fitted a logistic regression model using the "glm" function as shown below. library(rms) gusto <-

Hi all, I am fitting a linear mixed model with lme4 in R. The model has a single factor (des_days) with 4 levels (-1,1,14,48), and I am using random intercept and slopes. Fixed effects: data ~ des_days Value Std.Error DF t-value p-value (Intercept) 0.8274313 0.007937938 962 104.23757 0.0000 des_days1 -0.0026322 0.007443294 962 -0.35363 0.7237 des_days14 -0.0011319

Full_Name: Sahotra Sarkar Version: 2.2.0 OS: Windows XP Professional Submission from: (NULL) (146.6.130.180) The following two commands should give the same results for the eigenvectors but do not (there is a sign reversal for the first one): > summary(princomp(bumpus),loading = TRUE) Importance of components: Comp.1 Comp.2 Comp.3 Comp.4 Comp.5

Hi, the following strange error appears when I use qda: > qda1 <- qda(as.data.frame(mfilters[cvtrain,]),as.factor(traingroups)) Error: function is not a closure That's also strange: > qda1 <- qda(mfilters[cvtrain,],as.factor(traingroups)) Error in qda.default(mfilters[cvtrain, ], as.factor(traingroups)) : length of dimnames must match that of dims Some backgroud: >

rm(list=ls()) yx.df<-read.csv("c:/MK-2-72.csv",sep=',',header=T,dec='.') dim(yx.df) #get X matrix y<-yx.df[,1] x<-yx.df[,2:643] #conver to matrix mat<-as.matrix(x) #get row number rownum<-nrow(mat) #remove the constant parameters mat1<-mat[,apply(mat,2,function(.col)!(all(.col[1]==.col[2:rownum])))] dim(yx.df) dim(mat1) #remove columns with numbers of

Hi, In the following results I interpret exp(coef) as the factor that multiplies the base hazard rate if the corresponding variable is TRUE. For example, when the bucket is ks008 and fidelity <= 3, then the rate, compared to the base rate h_0(t), is h(t) = 0.200 h_0(t). My question is then, to what case does the base hazard rate correspond to? I would expect the reference to be the first